ESE319 Introduction to Microelectronics

Miller Effect BJT

2008 Kenneth R. Laker,update 18Oct10 KRL 1 ESE319 Introduction to Microelectronics

Prototype Circuit

Ignore “low frequency” High frequency model capacitors

2008 Kenneth R. Laker,update 18Oct10 KRL 2 ESE319 Introduction to Microelectronics

Multisim Simulation

Mid-band gain

∣Av∣=g m RC=40mS ∗5.1k =20446.2 dB

Half-gain point

2008 Kenneth R. Laker,update 18Oct10 KRL 3 ESE319 Introduction to Microelectronics Introducing the Miller Effect

The feedback connection of C  between base and collector causes it to appear to the amplifier like a large capacitor  1 − K  C  has been inserted between the base and emitter terminals. This phenomenon is called the “Miller effect” and the capacitive multiplier

“1 – K ” acting on C  equals the common emitter amplifier mid-band gain, i.e. K = − g m R C .

NOTE: and common collector do not suffer from the Miller effect, since in these amplifiers, one side of C  is connected directly to ground. 2008 Kenneth R. Laker,update 18Oct10 KRL 4 ESE319 Introduction to Microelectronics High Frequency CC and CB Models B C

E

ground B v-pi C B v-pi C

E E

C Common Collector Common Base C is in parallel with R . C is in parallel with R .  B  C

2008 Kenneth R. Laker,update 18Oct10 KRL 5 ESE319 Introduction to Microelectronics Miller's Theorem

I Z −I I 1=I I 2=−I + + + +

V 1 K V 1 V 2 V 1 Z Z V 2=K V 1 <=> 1 2 - - - -

V 1 V 1 Z V 1−V 2 V 1−K V 1 V 1 Z I = = = => 1= = = Z Z Z I 1 I 1−K 1−K 1 Z 1= 1 j 2 f C  1−K  V 2− V 2 V 2−V 2 K V 2 −I = = = V 2 V 2 Z Z Z Z Z = = = ≈ Z => 2 I −I 1 if K >> 1 1 2 1− 1− K K approx. V 2 V 2 1 Z = = ≈ open 2 I I j 2 f C 2 −   circuit

2008 Kenneth R. Laker,update 18Oct10 KRL 6 ESE319 Introduction to Microelectronics

Common Emitter Miller Effect Analysis

I Determine effect of C : C  B V C V Using phasor notation:  o I =−g V I I RC m  C  RC or

g V V o= −g m V I C RC m      where E I C = V −V o sC     

Note: The current through C  depends only on V  !

V o

2008 Kenneth R. Laker,update 18Oct10 KRL 7 ESE319 Introduction to Microelectronics Common Emitter Miller Effect Analysis II From slide 7:

I C = V g m V RC −I C RC s C       

Collect terms for I and V : C  

1s RC C I C = 1g m RC sC V       

1g R sC s 1g R C I  m C   V  m C   V C = =  1s RC C   1s RC C 

Miller Capacitance C : C eq=1−K C =1g m RC  eq 

2008 Kenneth R. Laker,update 18Oct10 KRL 8 ESE319 Introduction to Microelectronics

Common Emitter Miller Effect Analysis III

C eq=1g m RC  C 

For our example circuit:

1g m RC=10.040⋅5100=205

Ceq=205⋅2 pF≈410 pF

2008 Kenneth R. Laker,update 18Oct10 KRL 9 ESE319 Introduction to Microelectronics Apply Miller's Theorem to BJT CE Amplifier

−I I 2=−I

Z Z Z Z 2= 1= 1 1−K 1− K 1 For the BJT CE Amplifier: Z = and K=−gm RC j C  1 1 1 => Z 1= and Z 2= ≈ j 1g m RC C  1 j C  j 1 C  g m RC

C eq=1g m RC C  Miller's Theorem => important simplification to the HF BJT CE Model

2008 Kenneth R. Laker,update 18Oct10 KRL 10 ESE319 Introduction to Microelectronics Simplified HF Model

R C  s

C V R r V  r R R V s B  o C L o g m V 

' RL ' R =r ∥R ∥R ' I C L o C L  C  Rs

' RB∥r V V '' ' s= s V V C R V RB∥rRsig s   L o g m V '  Rs=r∥ RB∥Rs Thevenin

2008 Kenneth R. Laker,update 18Oct10 KRL 11 ESE319 Introduction to Microelectronics Simplified HF Model

' V s

Miller's Theorem

' I C  Rs

C in=C C eq  ' ' V V s C  C eq RL o ' .=C C  1g m RL g V C m  (c) in

2008 Kenneth R. Laker,update 18Oct10 KRL 12 ESE319 Introduction to Microelectronics Simplified HF Model Note that Z = 1/jωC 2 µ is ignored! Why?

' V s

V o dB V ∣ s∣ ' V o −gm RL Av  f = ≈ ' V s 1 j 2 f C in Rs 1 f H = ' 2C in Rs

2008 Kenneth R. Laker,update 18Oct10 KRL 13 ESE319 Introduction to Microelectronics

The Cascode Amplifier

● A two amplifier used to obtain simultaneously: 1. Reasonably high input impedance. 2. Reasonable voltage gain. 3. Wide bandwidth.

● None of the conventional single transistor designs will satisfy all of the criteria above. ● The cascode amplifier will satisfy all of these criteria. ● A cascode is a combination of a common emitter stage cascaded with a common base stage. (Historical Note: the cascode amplifier was a cascade of grounded cathode and grounded grid stages – hence the name “cascode,” which has remained in modern terminology.)

2008 Kenneth R. Laker,update 18Oct10 KRL 14 ESE319 Introduction to Microelectronics The Cascode Circuit CB Stage CE Stage CB Stage i R C1 R 1 C ic2 C byp i B1 C1 v-out vo B1 vO R ib2 s ie1 ic1

i E1 E1 CE Stage i R iC2 ie2 b1 R C 2i V CC v R s in B2 C2 s R C R E B2 B E2 i E2 v v v s R R R R R eg1 cg2 low 3 R B= 2∥ 3 in1= = = E ie1 ic2 ac equivalent circuit Comments:

1. R1, R2, R3, and RC set the bias levels for both Q1 and Q2.

2. Determine RE for the desired voltage gain.

3. Cin and Cbyp are to act as “open circuits” at dc and act as “short circuits” at all operating frequencies of interest, i.e. f  f min .

2008 Kenneth R. Laker,update 18Oct10 KRL 15 ESE319 Introduction to Microelectronics Cascode Mid-Band Small Signal Model

RB=R2∥R3

2008 Kenneth R. Laker,update 18Oct10 KRL 16 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis

ic1 1. Show reduction in Miller effect CB B1 vo 2. Evaluate small-signal voltage gain Stage C1 ib1 r v 1 g v 1  m 1 OBSERVATIONS E1 a. The emitter current of the CB stage is B2 ie1 R =low C2 ic2 in1 the collector current of the CE stage. (This R ib2 R s v g v C also holds for the dc bias current.) r  2 m 2 v 2 i i s E2 e1= c2 CE R ie2 R B E b. The base current of the CB stage is: Stage i i i = e1 = c2 b1 1 1 g m1=g m2=g m c. Hence, both stages have about same r r r e1= e2= e collector current i c1 ≈ i c2 and same g , r , r . m e π r1=r 2=r

2008 Kenneth R. Laker,update 18Oct10 KRL 17 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis cont. The input resistance R to the CB stage is in1 the small-signal “r ” for the CE stage, i.e. e1 v g v ie1 ic2 1 m 1 i = = b1 1 1 The CE output voltage, the voltage drop from Q2 collector to ground, is: v  2 g m v2 r 1 r 1 v =v =−r i =−  i =−  i c2 e1  1 b1 1 c2 1 e1 Therefore, the CB Stage input resistance is:

ve1 r1 Rin1= = =r e1 −ie1 1 v R r r A c2 in1 e 1 e vCE−Stage= ≈− =−  => C eq=1 C 2C  vs RE RE RE

2008 Kenneth R. Laker,update 18Oct10 KRL 18 ESE319 Introduction to Microelectronics Cascode Small Signal Analysis - cont. Now, find the CE collector current in terms of the input voltage v : i i s Recall c1≈ c2 v be11 g m v1 vs ib2≈ Rs∥RBr21 RE  v  v v s s vbe2 2 g m v2 ic2=ib2≈ ≈ Rs∥RBr 21 RE 1 RE

for bias insensitivity: 1RE ≫Rs∥RBr

v s vo −RC Av= = OBSERVATIONS: v s RE 1. Voltage gain A is about the same as a stand-along CE Amplifier. v 2. HF cutoff is much higher then a CE Amplifier due to the reduced C . eq 2008 Kenneth R. Laker,update 18Oct10 KRL 19 ESE319 Introduction to Microelectronics

Approximate Cascode HF Voltage Gain C µ2

v 1 g m v1 RC C − µ1 V o  f  RE Av  f = ≈ ' V s f  1 j 2 f C in Rs v  2 g m v2 where r C =C C =C 1 e C C 2C in  eq  R    ' E Rs=Rs∥RB≈Rs

2008 Kenneth R. Laker,update 18Oct10 KRL 20 ESE319 Introduction to Microelectronics Cascode Biasing

1. Choose IE1 – make it relatively large to

reduce R in 1 = r e1 = V T / I E1 to push out HF break frequencies. o.c. IC1 I 1 v C B O 2. Choose RC for suitable voltage swing R =low V and R for desired gain. I in1 C1 E E1 I C in C2 3. Choose bias resistor string such that I E2 its current I is about 0.1 of the collector o.c. 1 current I . C1

4. Given R , I and V = 0.7 V calc. R . 1 E E2 BE2 3 2 I E2=I C2=I E1= I C1 ⇒ I C1≈I E2 1 5. Need to also determine R & R . 1 2

2008 Kenneth R. Laker,update 18Oct10 KRL 21 ESE319 Introduction to Microelectronics Cascode Biasing - cont.

I I11 Since the CE-Stage gain is very small: v V O a. The collector swing of Q2 will be small. B1 b. The Q2 collector bias V = V - 0.7 V. C2 B1 V V 6. Set V V 1 V V 1V B2 C2 B1− B2= ⇒ CE2=

This will limit V V V V 0.3 V CB2 CB2= CE2− BE2= which will keep Q2 forward active.

7. Next determine R . Its drop V = 1 V V CE2=V C2−V R e=V C2−V B2−0.7V  2 R2 .=V −0.7V −V 0.7 V with the known current. V B1−V B2 B1 B2 R = 2 I .=V B1−V B2 1

2008 Kenneth R. Laker,update 18Oct10 KRL 22 ESE319 Introduction to Microelectronics Cascode Biasing - cont.

V B1−V B2 1V R2= = I 1 I 1 V 8. Then calculate R . R B2 3 3= I 1

where V B2=0.7V I E RE

V CC Note: R1R2R3= I 1 9. Then calculate R . 1

V CC R1= −R2−R3 0.1 I C

2008 Kenneth R. Laker,update 18Oct10 KRL 23 ESE319 Introduction to Microelectronics Cascode Bias Summary SPECIFIED: A , V , V (CB collector voltage); v CC C1 SPECIFIED: I (or I ) directly or indirectly through BW. E C DETERMINE: R , R , R , R and R . C E 1 2 3 V C1 RC STEP1: R RE= SET: V B1−V B2=1V ⇒V CE2≈1V C = I C ∣Av∣

V B1−V B2 1V STEP2: R2= = I 1 0.1 I C

V B2 0.7V I E RE STEP3: R3= = I 1 0.1 I C

V CC V CC R1R2R3= = I 1 0.1 I C V CC STEP4: R1= −R2−R3 0.1 I C

2008 Kenneth R. Laker,update 18Oct10 KRL 24 ESE319 Introduction to Microelectronics

Cascode Bias Example

RC I R VCC-ICRE-1.7 C C V C1 V V I R V I R VCE1CE1==ICCCRC−–1–IC CRCE− CE2− C E 1.0 =12 V =12 V VCE2=1

ICRE+0.7 RE ICRE

I E2≈I C2=I E1≈I C1 ⇒ I C1≈I E2 Cascode circuit Typical Bias Conditions

2008 Kenneth R. Laker,update 18Oct10 KRL 25 ESE319 Introduction to Microelectronics

Cascode Bias Example cont.

1. Choose IE1 – make it a bit high to lower re or r . Try I = 5 mA => r = 0.025 V / I = 5  .  E1 e E

V CE1=V CC−I C RC−1−I C RE

2. Set desired gain magnitude. For example if A = -10, then R /R = 10. V C E

3. Since the CE stage gain is very small, V can be small, i.e. V = V – V = 1 V. CE2 CE2 B1 B2

2008 Kenneth R. Laker,update 18Oct10 KRL 26 ESE319 Introduction to Microelectronics

Cascode Bias Example cont.

RC V CC=12 I C =5mA. ∣Av∣= =10 RE

V CE1=V CC−I C RC−1−I C RE

Determine RC for a 5 V drop across RC. 5V RC= =1000 5⋅10−3 A

RC RC R E= = =100 ∣Av∣ 10

2008 Kenneth R. Laker,update 18Oct10 KRL 27 ESE319 Introduction to Microelectronics

Cascode Bias Example cont.

I1 V CC=12 RC=1k  I C =5mA. RE=100

R1 R C I R V -I R -1.7 C C Make current through the string of bias CC C E resistors I = 1 mA. 1 VVCE1CE1==IVCRCCC−–1–II CCRRCE −1−I C RE R V CC 12 2 1.0 R R R 12k 1 2 3= = −3 =  =12V I 1 1⋅10 VCE2=1 Calculate the bias voltages (base side of Q1, Q2): ICRE+0.7 R3 R E I R C E V CC −I C R E−1.7V =12V −0.5V −1.7V =9.8V

V B1B2=V B1−V B2=1.0V −3 V B2=I C RE0.7=5⋅10 ⋅1000.7=1.2V

2008 Kenneth R. Laker,update 18Oct10 KRL 28 ESE319 Introduction to Microelectronics

Cascode Bias Example cont.

−3 V B2=10 R3=1.2V

C B V B1 R3=1.2 k  −3 C in V B1−V B2=1⋅10 R2=1.0V V B2 R2=1 k 

Recall: R1R2R3=12k 

R1=12000−1200−1000=9.8k 

R1=10 k  V CC=12 RC=1k  V B2=1.2V

I C =5mA. RE=100 V B1−V B2=1.0V

2008 Kenneth R. Laker,update 18Oct10 KRL 29 ESE319 Introduction to Microelectronics Multisim Results – Bias Example I R C1 R 1 C vo C B

R R 2 s C in

v R s 3 R E

V CC −7.3290.9710.339 12V −8.639 V Check I : I 3.36 mA C1 C1= = = RC 1000 IC1 about 3.4 mA. That's a little low.

Increase R3 to 1.5 k Ohms and re-simulate. 2008 Kenneth R. Laker,update 18Oct10 KRL 30 ESE319 Introduction to Microelectronics Improved Biasing

R R C 1

C B

10 uF R R 2 s C in

10 uF v R s 3 R E

V CC −5.0480.9410.551 12V −6.540 V Check I : I 5.46mA C1 C1= = = RC 1000 That's better! Now measure the gain at a mid-band frequency with some “large” coupling capacitors, say 10 µF inserted. 2008 Kenneth R. Laker,update 18Oct10 KRL 31 ESE319 Introduction to Microelectronics Single Frequency Gain

R R 1 C 10 uF v-outv C B o

R R 2 s 10 uF C in

v s R 3 R E

Gain |A | of about 8.75 at 100 kHz - OK for rough calculations. Some attenua- v tion from low CB input impedance (RB = R2||R3)and some from 5Ω re.

2008 Kenneth R. Laker,update 18Oct10 KRL 32 ESE319 Introduction to Microelectronics Bode Plot for the Amplifier

18.84 dB

Low frequency break point with 10 µF. capacitors

18.84 dB

High frequency break point – internal capacitances only

2008 Kenneth R. Laker,update 18Oct10 KRL 33 ESE319 Introduction to Microelectronics Scope Plot – Near 5 V Swing on Output

2008 Kenneth R. Laker,update 18Oct10 KRL 34 ESE319 Introduction to Microelectronics

Determine Bypass Capacitors Low Frequency f ≤ f min C v-out B From CE stage determine C in 10 10 C in C in ≥ ≈ F 2 f min RB∥r bg 2 f min RB∥1 RE

RB=R2∥R3

From CB stage determine C B 10 C ≥ F B 2 f 1 R r  min   E  where R -> R S E

2008 Kenneth R. Laker,update 18Oct10 KRL 35